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A newspaper published an article about a study in which researchers subjected laboratory gloves to stress. Among 215 vinyl glov

pickupchik [31]

Answer:

$z=\frac{0.7-0.07}{\sqrt{0.385(1-0.385)(\frac{1}{215}+\frac{1}{215})}}=13.42$

$p_v =P(Z>13.42)\approx 0$

So the p value is a very low value and using any significance level given $\alpha=0.01$ we se that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that vinyl gloves have a greater virus leak rate than latex gloves at 1% of significance.

Step-by-step explanation:

Data given and notation

$n_{V}=215$ sample of vinyl gloves selected

$n_{L}=215$ sample of latex gloves selected

$p_{V}=0.7$ represent the proportion of vynil gloves with leaked viruses

$p_{L}=0.07$ represent the proportion of latex gloves with leaked viruses

z would represent the statistic (variable of interest)

$p_v$ represent the value for the test (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to check if the vinyl gloves have a greater virus leak rate than latex gloves , the system of hypothesis would be:

Null hypothesis: $p_{V} \leq p_{L}$

Alternative hypothesis: $p_{V} > \mu_{L}$

We need to apply a z test to compare proportions, and the statistic is given by:

$z=\frac{p_{V}-p_{L}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{V}}+\frac{1}{n_{L}})}}$ (1)

Where $\hat p=\frac{X_{V}+X_{L}}{n_{V}+n_{L}}=\frac{0.7+0.07}{2}=0.385$

Calculate the statistic

Replacing in formula (1) the values obtained we got this:

$z=\frac{0.7-0.07}{\sqrt{0.385(1-0.385)(\frac{1}{215}+\frac{1}{215})}}=13.42$

Statistical decision

Since is a one right tailed test the p value would be:

$p_v =P(Z>13.42)\approx 0$

So the p value is a very low value and using any significance level given $\alpha=0.01$ we se that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that vinyl gloves have a greater virus leak rate than latex gloves at 1% of significance.

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